LOJ #6278. 数列分块入门 2
思路&&代码
区间修改+询问区间内小于某个值x的元素个数
还是先分块,分成(sqrt{n})个块,分的时候,每个块用一个(vector)来维护块内的值
每个块内排序,保证块内是有序的(便于用(lower\_bound)统计答案),因为分好了这个块以后如果整体修改不会对块内元素造成影响,如果不是整个块修改的话,可以修改之后再顺便排序,覆盖之前这个块的值
统计答案的时候整个块用(lower\_bound),局部直接暴力枚举,还是(O(nsqrt{n}))
#include <cmath>
#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int A = 1e5 + 11;
inline int read() {
char c = getchar(); int x = 0, f = 1;
for ( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
for ( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
return x * f;
}
vector <int> v[500];
int n, m, a[A], L[A], R[A], add[A], pos[A];
void reset(int x) {
v[x].clear();
for(int i = L[x]; i <= R[x]; i++) v[x].push_back(a[i]);
sort(v[x].begin(), v[x].end());
}
void change(int l, int r, int c) {
int p = pos[l], q = pos[r];
if(p == q) {
for(int i = l; i <= r; i++) a[i] += c;
reset(p); return;
}
for(int i = l; i <= R[p]; i++) a[i] += c; reset(p);
for(int i = L[q]; i <= r; i++) a[i] += c; reset(q);
for(int i = p + 1; i < q; i++) add[i] += c;
}
int ask(int l, int r, int c, int ans = 0) {
int p = pos[l], q = pos[r];
if(p == q) {
for(int i = l; i <= r; i++) if(a[i] + add[p] < c) ans++;
return ans;
}
for(int i = l; i <= R[p]; i++) if(a[i] + add[p] < c) ans++;
for(int i = L[q]; i <= r; i++) if(a[i] + add[q] < c) ans++;
for(int i = p + 1; i < q; i++) ans += lower_bound(v[i].begin(), v[i].end(), c - add[i]) - v[i].begin();
return ans;
}
int main() {
n = read();
for(int i = 1; i <= n; i++) a[i] = read();
int t = sqrt(n);
for(int i = 1; i <= t; i++) {
L[i] = sqrt(n) * (i - 1) + 1;
R[i] = sqrt(n) * i;
}
if(R[t] < n) t++, L[t] = R[t - 1] + 1, R[t] = n;
for(int i = 1; i <= t; i++) {
for(int j = L[i]; j <= R[i]; j++) {
pos[j] = i;
v[pos[j]].push_back(a[j]);
}
}
for(int i = 1; i <= t; i++) sort(v[i].begin(), v[i].end());
m = n;
while(m--) {
int opt = read(), l = read(), r = read(), c = read();
if(opt == 0) change(l, r, c);
else cout << ask(l, r, c * c) << '
';
}
return 0;
}